如果您真的想避免使用它，那么唯一的选择是元类或单独的factory函数/方法^{(但是请参阅下面的Python3.6+更新)}。如果有可用的选择，使__new__()方法成为一个方法--因为它在默认情况下是静态的--是非常明智的方法。

也就是说，下面是我认为的代码的改进版本。我已经添加了几个类方法来帮助自动查找所有子类。它们支持最重要的更好的方式--现在添加子类不需要修改__new__()方法。这意味着它现在很容易扩展，因为它有效地支持您所说的虚拟构造函数。

还可以使用类似的实现将实例的创建从__new__()方法移到单独的(静态)工厂方法中-因此，从某种意义上讲，所示的技术只是对可扩展的泛型工厂函数进行编码的一种相对简单的方式，而不管它被赋予什么名称。

# Works in Python 2 and 3.

import os
import re

class FileSystem(object):
    class NoAccess(Exception): pass
    class Unknown(Exception): pass

    # Regex for matching "xxx://" where x is any non-whitespace character except for ":".
    _PATH_PREFIX_PATTERN = re.compile(r's*([^:]+)://')

    @classmethod
    def _get_all_subclasses(cls):
        """ Recursive generator of all class' subclasses. """
        for subclass in cls.__subclasses__():
            yield subclass
            for subclass in subclass._get_all_subclasses():
                yield subclass

    @classmethod
    def _get_prefix(cls, s):
        """ Extract any file system prefix at beginning of string s and
            return a lowercase version of it or None when there isn't one.
        """
        match = cls._PATH_PREFIX_PATTERN.match(s)
        return match.group(1).lower() if match else None

    def __new__(cls, path):
        """ Create instance of appropriate subclass using path prefix. """
        path_prefix = cls._get_prefix(path)

        for subclass in cls._get_all_subclasses():
            if subclass.prefix == path_prefix:
                # Using "object" base class method avoids recursion here.
                return object.__new__(subclass)
        else:  # No subclass with matching prefix found (& no default defined)
            raise FileSystem.Unknown(
                'path "{}" has no known file system prefix'.format(path))

    def count_files(self):
        raise NotImplementedError


class Nfs(FileSystem):
    prefix = 'nfs'

    def __init__ (self, path):
        pass

    def count_files(self):
        pass


class LocalDrive(FileSystem):
    prefix = None  # Default when no file system prefix is found.

    def __init__(self, path):
        if not os.access(path, os.R_OK):
            raise FileSystem.NoAccess('Cannot read directory')
        self.path = path

    def count_files(self):
        return sum(os.path.isfile(os.path.join(self.path, filename))
                     for filename in os.listdir(self.path))


if __name__ == '__main__':

    data1 = FileSystem('nfs://192.168.1.18')
    data2 = FileSystem('c:/')  # Change as necessary for testing.

    print(type(data1).__name__)  # -> Nfs
    print(type(data2).__name__)  # -> LocalDrive

    print(data2.count_files())  # -> <some number>

Python3.6+更新

我从PEP 487 -- Simpler customisation of class creation提案中的Subclass registration部分得到了使用__init_subclass__()来完成此操作的想法。由于该方法将由所有基类的子类继承，因此子类的注册也将自动完成(与仅直接子类相反)-它完全不需要像_get_all_subclasses()这样的方法。

# Requires Python 3.6+

import os
import re

class FileSystem(object):
    class NoAccess(Exception): pass
    class Unknown(Exception): pass

    # Pattern for matching "xxx://"  # x is any non-whitespace character except for ":".
    _PATH_PREFIX_PATTERN = re.compile(r's*([^:]+)://')
    _registry = {}  # Registered subclasses.

    @classmethod
    def __init_subclass__(cls, /, path_prefix, **kwargs):
        super().__init_subclass__(**kwargs)
        cls._registry[path_prefix] = cls  # Add class to registry.

    @classmethod
    def _get_prefix(cls, s):
        """ Extract any file system prefix at beginning of string s and
            return a lowercase version of it or None when there isn't one.
        """
        match = cls._PATH_PREFIX_PATTERN.match(s)
        return match.group(1).lower() if match else None

    def __new__(cls, path):
        """ Create instance of appropriate subclass. """
        path_prefix = cls._get_prefix(path)
        subclass = cls._registry.get(path_prefix)
        if subclass:
            return object.__new__(subclass)
        else:  # No subclass with matching prefix found (and no default).
            raise cls.Unknown(
                f'path "{path}" has no known file system prefix')

    def count_files(self):
        raise NotImplementedError


class Nfs(FileSystem, path_prefix='nfs'):
    def __init__ (self, path):
        pass

    def count_files(self):
        pass

class Ufs(Nfs, path_prefix='ufs'):
    def __init__ (self, path):
        pass

    def count_files(self):
        pass

class LocalDrive(FileSystem, path_prefix=None):  # Default file system.
    def __init__(self, path):
        if not os.access(path, os.R_OK):
            raise self.NoAccess(f'Cannot read directory {path!r}')
        self.path = path

    def count_files(self):
        return sum(os.path.isfile(os.path.join(self.path, filename))
                     for filename in os.listdir(self.path))


if __name__ == '__main__':

    data1 = FileSystem('nfs://192.168.1.18')
    data2 = FileSystem('c:/')  # Change as necessary for testing.
    data4 = FileSystem('ufs://192.168.1.18')

    print(type(data1))  # -> <class '__main__.Nfs'>
    print(type(data2))  # -> <class '__main__.LocalDrive'>
    print(f'file count: {data2.count_files()}')  # -> file count: <some number>

    try:
        data3 = FileSystem('c:/foobar')  # A non-existent directory.
    except FileSystem.NoAccess as exc:
        print(f'{exc} - FileSystem.NoAccess exception raised as expected')
    else:
        raise RuntimeError("Non-existent path should have raised Exception!")

    try:
        data4 = FileSystem('foobar://42')  # Unregistered path prefix.
    except FileSystem.Unknown as exc:
        print(f'{exc} - FileSystem.Unknown exception raised as expected')
    else:
        raise RuntimeError("Unregistered path prefix should have raised Exception!")

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